The modular group and its action on the upper half-plane

Let

$S{L}_2(\mathbb{Z})=\{\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right):a,b,c,d\in \mathbb{Z},ad-bc=1\},$

a subgroup of $G{L}_2(\mathbb{R})\subseteq G{L}_2(\mathbb{C})$ (Invertible matrices and the general linear group ${\mathrm{GL}}_n(F)$, ${\mathrm{GL}}_n(F)$ is a group under matrix multiplication, including the trivial group ${\mathrm{GL}}_0(F)$, The integers as equivalence classes of pairs of naturals, For same-sized finite square matrices over a commutative ring, $\det (AB)=\det (A)\det (B)$). Its centre is $\{\pm I\}$: the inclusion $\{\pm I\}\subseteq Z(S{L}_2(\mathbb{Z}))$ is immediate, $\{\pm I\}$ is normal (The center of a group is a normal subgroup, Normal subgroup: invariance under conjugation), and conversely every $2\times 2$ matrix commuting with all of $S{L}_2(\mathbb{Z})$ is scalar — in particular a matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ commuting with both $U=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ and $L=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ satisfies $c=0$ and $a=d$ from $AU=UA$, and then $b=0$ from $AL=LA$, by comparing entries of the two products, so $A=aI$ is scalar; a scalar matrix $\lambda I$ of determinant $1$ has ${\lambda }^2=1$, so $\lambda =\pm 1$ (Rectangular matrix multiplication and the identity matrix $I_n$, including zero-sized shapes). The modular group is the quotient

$PS{L}_2(\mathbb{Z}):=S{L}_2(\mathbb{Z})\mathrm{/}\{\pm I\},$

a group by For $N⊴G$, the cosets form a group with identity $N$ and inverse $(gN{)}^{-1}=g^{-1}N$, The quotient group $G\mathrm{/}N$ and coset product $(gN)(hN)=ghN$, The canonical projection $\pi :G\to G\mathrm{/}N$, $\pi (g)=gN$, is a surjective group homomorphism.

For $\gamma \in S{L}_2(\mathbb{R})$ and $\tau \in \mathfrak{H}$ set $\gamma \cdot \tau :=\frac{a\tau +b}{c\tau +d}$, the Möbius transformation attached to $\gamma$ (Möbius transformations of the Riemann sphere, The unit disc, the upper half-plane, and Blaschke factors). The denominator does not vanish: $c\tau +d=0$ would give $\tau =-d\mathrm{/}c\in \mathbb{R}$ when $c\mathrm{\ne }0$, and when $c=0$ invertibility gives $d\mathrm{\ne }0$. Writing $c\tau +d=(c\mathrm{Re}\tau +d)+ic\mathrm{Im}\tau$ and using $\overline{c\tau +d}=c\bar{\tau }+d$, a direct computation gives

$\mathrm{Im}(\gamma \cdot \tau )=\frac{\mathrm{Im}\tau }{\mathrm{\mid }c\tau +d{\mathrm{\mid }}^2}>0,$

so $\mathfrak{H}$ is stable under each $\gamma$, and $\gamma \mapsto \gamma \cdot$ is the restriction of the matrix-to-Möbius map, a group homomorphism (Möbius transformations form a group and identify with the projective linear quotient of GL_2(C), Actions of $G$ on $X$ correspond exactly to homomorphisms $G\to \mathrm{Sym}(X)$); hence $\gamma \mapsto (\tau \mapsto \gamma \cdot \tau )$ is a left action of $S{L}_2(\mathbb{R})$ on $\mathfrak{H}$ by biholomorphisms (Left group actions, transitive actions, and faithful actions, Every Möbius transformation is a biholomorphism of the Riemann sphere, Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions, Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive, Real and imaginary parts, complex conjugation, and modulus)).

The kernel of the restricted action of $S{L}_2(\mathbb{Z})$ is exactly $\{\pm I\}$: $\pm I$ act trivially, while a matrix acting trivially fixes the three distinct points $i$, $i+1$, $2i\in \mathfrak{H}$, so it is the identity Möbius transformation and hence scalar — knowing that a Möbius transformation is determined by its values at three distinct points and that the kernel of the matrix-to-Möbius map is the scalar subgroup (A unique Möbius transformation carries any ordered triple of distinct sphere points to any other, Möbius transformations form a group and identify with the projective linear quotient of GL_2(C)) — and a scalar in $S{L}_2(\mathbb{Z})$ is $\pm I$. The action therefore descends to a faithful action of $PS{L}_2(\mathbb{Z})$, the quotient by the normal subgroup $\{\pm I\}$ (First isomorphism theorem for groups: $G\mathrm{/}\ker f\cong \mathrm{im}f$).

The elements $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ act by $S\cdot \tau =-1\mathrm{/}\tau$ and $T\cdot \tau =\tau +1$. In $S{L}_2(\mathbb{Z})$ one computes $S^2=-I$ and $(ST{)}^3=-I$: $ST=\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ has $(ST{)}^2-ST+I=0$ by direct multiplication, whence $(ST{)}^3=(ST)(ST-I)=-I$. Hence $S$ and $ST$ have order $2$ and $3$ in $PS{L}_2(\mathbb{Z})$, so $S^2=1$ and $(ST{)}^3=1$ there.